Taking into account the scientific notation, the result of the sum is 10.84300×10³.
- <u><em>Scientific notation</em></u>
First, remember that scientific notation is a quick way to represent a number using powers of base ten.
The numbers are written as a product:
a×10ⁿ
where:
- a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
- n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.
-
<u><em>Sum in scientific notation</em></u>
You want to add two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found. In this case, the highest exponent is 3.
Then all the values are expressed as a function of the base 10 exponent with the highest exponent. In this case: 9.7300×10²= 0.97300×10³
Taking the quantities to the same exponent, all you have to do is add what was previously called the number "a". In this case:
0.97300×10³ + 9.8700×10³= (0.97300+ 9.8700)×10³= 10.84300×10³
Finally, the result of the sum is 10.84300×10³.
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Answer: 41.5 mL
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,
n = moles of solute
= volume of solution in L
Given : 59.4 g of
in 100 g of solution
moles of 
Volume of solution =
Now put all the given values in the formula of molality, we get

To calculate the volume of acid, we use the equation given by neutralisation reaction:

where,
are the molarity and volume of stock acid which is 
are the molarity and volume of dilute acid which is 
We are given:

Putting values in above equation, we get:

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid
O valence electron number is the answer
B boiling point https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/03%3A_Matter_and_Energy/3.05%3A_Differences_in_Matter%3A_Physical_and_Chemical_Properties#Summary
Answer:
94.325 g
Explanation:
We'll begin by converting 350 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
350 mL = 350 mL × 1 L /1000 mL
350 mL = 0.35 L
Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:
Volume = 0.35 L
Molarity of KC₂H₃O₂ = 2.75 M
Mole of KC₂H₃O₂ =?
Molarity = mole /Volume
2.75 = Mole of KC₂H₃O₂ / 0.35
Cross multiply
Mole of KC₂H₃O₂ = 2.75 × 0.35
Mole of KC₂H₃O₂ = 0.9625 mole
Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:
Mole of KC₂H₃O₂ = 0.9625 mole
Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)
= 39 + 24 + 3 + 32
= 98 g/mol
Mass of KC₂H₃O₂ =?
Mass = mole × molar mass
Mass of KC₂H₃O₂ = 0.9625 × 98
Mass of KC₂H₃O₂ = 94.325 g
Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g